Junior problems
J79. Find all integers that can be represented as a3+ b3+ c3− 3abc for some positive integers a, b, and c.
Proposed by Titu Andreescu, University of Texas at Dallas, USA
First solution by Salem Malikic Sarajevo, Bosnia and Herzegovina We have
w = a3+ b3+ c3− 3abc = (a + b + c)(a2+ b2+ c2− ab − bc − ca) = = (a + b + c)
2 [(a − b)
2+ (b − c)2+ (c − a)2].
Let a = b = n + 1 and c = n, then w = 3n + 2 so each positive integer of the form 3n + 2 can be written as a3+ b3+ c3− 3abc. Taking a = n + 1, b = c = n we can represent all positive integers of the form 3n + 1.
If we take a = n − 1, b = n, and c = n + 1 we find that w = 9n, n ≥ 2. Thus each positive integer divisible by 9 except 9 can be written in the given form. Now we will prove that integers of the form 9k + 3 and 9k + 6 cannot be written as
(a + b + c)
2 [(a − b)
2+ (b − c)2+ (c − a)2].
Since x2 ≡ 0, 1 (mod 3) the number (a − b)2+ (b − c)2+ (c − a)2 is divisible by
three if and only if (a − b)2 ≡ (b − c)2 ≡ (c − a)2 ≡ 0 (mod 3) or (a − b)2 ≡
(b − c)2 ≡ (c − a)2≡ 1 (mod 3).
If (a − b)2 ≡ (b − c)2≡ (c − a)2≡ 0 (mod 3) then a ≡ b ≡ c ≡ 0 (mod 3). Thus
3|(a + b + c) so 9|12(a + b + c)[(a − b)2+ (b − c)2+ (c − a)2] and so w 6= 9k ± 3. If (a−b)2 ≡ (b−c)2≡ (c−a)2≡ 1 (mod 3) then no two of a, b, c give same residue
mod 3 and we can assume without loss of generality that a ≡ 1, b ≡ 2, and c ≡ 0 (mod 3) and so a+b+c ≡ 0 (mod 3) and 9 | 12(a+b+c)[(a−b)2+(b−c)2+(c−a)2] which means w 6= 9k ± 3. From these, we conclude that, if w is of the form 9k ± 3 then 3 | a + b + c, while 3 does not divide (a − b)2+ (b − c)2+ (c − a)2.
It is easy to check that 3 | a + b + c if one of the following residue situation occurs:
(a, b, c) = (0, 0, 0), (1, 1, 1), (2, 2, 2), (2, 1, 0).
In each case (a − b)2+ (b − c)2+ (c − a)2 is divisible by three so if 12(a + b + c)[(a − b)2+ (b − c)2+ (c − a)2] is divisible by 3, then it is divisible by 9 and therefore it cannot be of the form 9k ± 3.
Second solution by Daniel Campos Salas, Costa Rica
Let us say an integer is nice if it can be represented as a3+ b3+ c3− 3abc for some positive integers a, b, c. Assume without loss of generality that b = a + x and c = a + x + y, for some nonnegative integers x, y. Therefore,
a3+ b3+ c3− 3abc = (3a + 2x + y)(x2+ xy + y2).
For x = y = 0 it follows that 0 is nice. Suppose that x, y are not both zero. Since (3a + 2x + y)(x2+ xy + y2) > 0 we have that any nonzero nice integer is nonnegative. Let us prove first that 1 and 2 are not nice. We have that
(3a + 2x + y)(x2+ xy + y2) > 3a(x2+ xy + y2) ≥ 3,
from where it follows the claim. Let us prove also that any nice integer divisible by 3 must be divisible by 9. We have that
0 ≡ (3a + 2x + y)(x2+ xy + y2) ≡ (y − x) · (x − y)2 ≡ (y − x)3 (mod 3),
from where it follows that x ≡ y (mod 3). Therefore, 3a + 2x + y ≡ x2+ xy + y2 ≡ 0 (mod 3),
which implies the claim. Let us prove that 9 is not nice. From the previous result we have that x ≡ y (mod 3), from where it follows that
(3a + 2x + y)(x2+ xy + y2) ≥ (3a + 3) · 3 > 9.
Let us proceed to find which integers are nice. Taking x = 0, y = 1 it follows that any positive integer of the form 3a + 1 is nice. Taking x = 1, y = 0 it follows that any positive integer of the form 3a + 2 is nice. Taking x = y = 1 it follows that any positive integer of the form 9(a + 1) is nice. From these we conclude that all the nice integers are 0, any positive integer greater than 3 of the form 3a + 1 or 3a + 2, and the integers greater than 9 of the form 9a, and we are done.
Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; John T. Robinson, Yorktown Heights, NY, USA; Roberto Bosch Cabrera, Faculty of Mathematics, University of Havana, Cuba.
J80. Characterize triangles with sidelengths in arithmetical progression and lengths of medians also in arithmetical progression.
Proposed by Daniel Lasaosa, Universidad Publica de Navarra, Spain
First solution by Magkos Athanasios, Kozani, Greece
We prove that only equilateral triangles have the desired property. Let ABC be a triangle with sides a, b, c and the corresponding medians ma, mb, mc. Assume
that a ≥ b ≥ c. Then we have ma≤ mb ≤ mc. Since the sides and the medians
form arithmetic progressions we have
2b = a + c, 2mb= ma+ mc. (1)
It is a known fact that m2a+ m2b+ m2c = 34(a2+ b2+ c2). and we also know that for reals x, y, z we have the inequality 3(x2+ y2+ z2) ≥ (x + y + z)2. Hence, we have 3 4(a 2+ b2+ c2) = m2 a+ m2b + m2c ≥ 1 3(ma+ mb+ mc) 2.
From this, using (1) and the relation 4m2
b = 2a2 + 2c2 − b2, we obtain the
following chain of inequalities
9 4(a
2+ b2+ c2) ≥ (3m
b)2 ⇔ a2+ b2+ c2 ≥ 4m2b ⇔ a2+ b2+ c2≥ 2a2+ 2c2−b2.
Hence, 2b2 ≥ a2 + c2 ⇔ (a + c)2 ≥ 2a2+ 2c2 ⇔ (a − c)2 ≤ 0. This means
that a = c. Therefore, we have a = b = c.
Second solution by Vicente Vicario Garcia, Huelva, Spain
We use the habitual notation in a triangle. Without loss of generality a ≤ b ≤ c. By the well known Apollonius-formulas (application of Stewart’s theorem) for the medians of a triangle we have
mA=
1 2
p
2b2+ 2c2− a2
and the analoguous ones. We can then deduce that mC ≤ mB ≤ mA. By the
properties of arithmetic progression we have that a + c = 2b [1] mA+ mC = 2mB [2].
By the relations [1] and [2] we have p
2b2+ 2c2− a2+p2a2+ 2b2− c2 = 2p2a2+ 2c2− b2 ⇔
a2+ 4b2+ 2p(2b2+ 2c2− a2)(2a2+ 2b2− c2) = 4(2a2+ 2c2− b2).
Doing all the subsequent calculations will yeild 5c a 4 − 8c a 3 + 6c a 2 − 8c a + 5 = 0.
Then the polynomial P (x) = 5x4−8x3+6x2−8x+5 = (x−1)2(5x2+2x+5) and
the quadratic equation 5x2+ 2x + 5 has no real roots. Then c
a = 1 which means
a = b = c and the triangle is equilateral. Finally, it is clear that equilateral triangle satisfies the problem and we are done.
Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; John T. Robinson, Yorktown Heights, NY, USA; Roberto Bosch Cabrera, Faculty of Mathematics, University of Havana, Cuba.
J81. Let a, b, c be positive real numbers such that 1 a2+ b2+ 1+ 1 b2+ c2+ 1+ 1 c2+ a2+ 1 ≥ 1. Prove that ab + bc + ca ≤ 3.
Proposed by Alex Anderson, New Trier High School, Winnetka, USA
First solution by Dinh Cao Phan, Pleiku, GiaLai, Vietnam By applying the Cauchy-Scharz inequlity, we have
(a2+ b2+ 1)(1 + 1 + c2) ≥ (a + b + c)2 or 1 a2+ b2+ 1 ≤ 2 + c2 (a + b + c)2. Similarily, we obtain 1 b2+ c2+ 1≤ 2 + a2 (a + b + c)2, 1 c2+ a2+ 1≤ 2 + b2 (a + b + c)2. Therefore 1 ≤ 1 a2+ b2+ 1+ 1 b2+ c2+ 1+ 1 c2+ a2+ 1 ≤ 6 + a2+ b2+ c2 (a + b + c)2 or (a + b + c)2≤ 6 + a2+ b2+ c2 equivalent to a2+ b2+ c2+ 2(ab + bc + ca) ≤ 6 + a2+ b2+ c2. Thus ab + bc + ca ≤ 3 and we are done.
Also solved by Oleh Faynshteyn, Leipzig, Germany; Daniel Campos Salas, Costa Rica; Daniel Lasaosa, Universidad Publica de Navarra, Spain; Nguyen Manh Dung, Hanoi, Vietnam; Perfetti Paolo, Dipartimento di matematica Universita degli studi di Tor Vergata, Italy.
J82. Let ABCD be a quadrilateral whose diagonals are perpendicular. Denote by Ω1, Ω2, Ω3, Ω4 the centers of the nine-point circles of triangles ABC, BCD,
CDA, DAB, respectively. Prove that the diagonals of Ω1Ω2Ω3Ω4 intersect at
the centroid of ABCD.
Proposed by Ivan Borsenco, University of Texas at Dallas, USA
First solution by G.R.A.20 Math Problems Group, Roma, Italy
Let MAB, MBC, MCD, MAD be the midpoints of the sides AB, BC, CD, DA,
respectively. Since the nine-point circle of the triangle ABC passes through the midpoints of its sides we have Ω1 belongs to the perpedicular bisector of
MABMBC. Similarly, Ω3 belongs to the perpedicular bisector of MCDMDA.
Since AC and BD are perpendicular we get that MABMBCMCDMAD is a
rectangle. This implies that the line Ω1Ω3 is the midline of opposite sides of
this rectangle: MABMBC and MCDMDA. Finally, the intersection of the lines
Ω1Ω3 and Ω2Ω4 coincides with the intersection of the diagonals of the rectangle
MABMBCMCDMAD which is the centroid of ABCD.
Second solution by Roberto Bosch Cabrera, University of Havana, Cuba
We proceed by coordinate geometry. Let the point of intersection of the diag-onals be (0, 0). Let A = (a, 0), B = (0, b), C = (c, 0), D = (0, d), then the centroid of ABCD, G = (a+c4 ,b+d4 ).
Now we will find the coordinates of Ω1. Let H and O be the orthocenter
and the circumcenter of triangle ABC. Then H = (0,−acb ) and using that BH = 2OF where F is the feet of the perpendicular to AC from O we obtain that O = (a+c2 ,b22b+ac). It is well known that Ω1 is the midpoint of HO, hence
Ω1 = (a+c4 ,b 2−ac 4b ). Analogously, Ω2 = (c 2−bd 4c , b+d 4 ), Ω3 = ( a+c 4 , d2−ac 4d ), Ω4 = ( a2−bd 4a , b+d 4 ). Clearly,
Ω1Ω3 and Ω2Ω4 pass through G, and we are done.
Third solution by Mihai Miculita, Oradea, Romania
Let Macand Mbdbe the midpoints of the diagonals AC and BD. It is known that
the centroid G of ABCD coincides with the midpoint of MacMbd. Let Ox and
Hx be the cicrumcenter and the orthocenter of triangle Y ZT , {X, Y, Z, T } =
{A, B, C, D}. Points Odand Oc being cicrumcenters of triangle ABC and ACD,
are on the same perpendicular bisector of AC, yielding Mac ∈ OdOb⊥AC. (1)
The quadrilateral ABCD having the diagonals intersecting at a right angle implies that BO and DO are heights in the triangles ABC and ACD. Thus Hd, Hb ∈ BD and
HdHB⊥AC. (2).
From (1) and (2) it follows that OdOb k HdHb. Last relation proves that
OdObHdHb is a trapezoid. The Euler circle’s center in a triangle is the
mid-point of the segment determined by the circumcenter and the orthocenter. It follows that Ωd and Ωb are the midpoints of OdHd and ObHb. Thus the line
ΩdΩb is the midline of this trapezoid and so passes through G, the midpoint of
MacMbd. Analogoulsly, we prove that G ∈ ΩaΩc.
Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; Vicente Vicario Garcia, Huelva, Spain.
J83. Find all positive integers n such that a divides n for all odd positive integers a not exceeding√n.
Proposed by Dorin Andrica, Babes-Bolyai University, Romania
First solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain; If 1 is the largest odd integer not exceeding√n, the result is trivially true, and √
n < 3, or n ≤ 8. Assume now that m ≥ 1 is an integer such that 2m + 1 is the largest odd integer not exceeding √n. Then, 2m + 3 > √n ≥ 2m + 1, or 4m2+ 12m + 9 > n ≥ 4m2+ 4m + 1. Since 2m + 1 and 2m − 1 are positive odd integers with difference 2, they are coprime, and if both divide n, then their product 4m2− 1 must also divide n, which is larger than 4m2− 1. Therefore,
n ≥ 2(4m2 − 1), and 4m2 + 12m + 9 > 8m2 − 2, or 4m2 − 12m − 11 < 0.
Now, if m ≥ 4, then 4m2 − 12m − 11 = (m2 − 11) + 3(m − 4)m > 0, and
necessarily m ≤ 3. Assume that m = 3. Then 2m + 3 = 9 >√n, and n < 81, but n must be divisible by 3, 5 and 7, which are coprime. Therefore, n must be divisible by 105, which is absurd, and m ≤ 2. Assume now m = 2. Then, 2m + 3 = 7 >√n ≥ 5 = 2m + 1, and 25 ≤ n < 49, but n must be divisible by 3 and 5, which are coprime. Therefore, n must be divisible by 15, or n = 30, 45. Assume next that m = 1. Then, 9 ≤ n < 25 and n must be divisible by 3, or n = 9, 12, 15, 18, 21, 24. The integers that we are looking for are then 1 through 9, 12, 15, 18, 21, 24, 30 and 45.
Second solution by John T. Robinson, Yorktown Heights, NY, USA
For n ≤ 25 such integers can be computed as 1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, and 24. In order to get some intuition, consider what happens for n ≥ 25: up to n = 72 = 49, these are the integers divisible by 3 and 5, that is multiples of 15, which are 30 and 45 in this range. After n = 49, up to n = 92= 81, these are the integers divisible by 3, 5, and 7, that is multiples of 105, a contradiction. Recall Bertrand’s postulate, that is there is always a prime between m and 2m, where m is any integer with m > 1. Using induction we can see that every time we jump from n to 4n we get at least one more prime in the range [√n, 2√n]. This prime is greater than 4 for n ≥ 9, so the product of primes that must divide n grows faster than n. In summary, the only positive integers n such that a divides n for all odd positive integers a not exceeding √n are
Third solution by Roberto Bosch Cabrera, University of Havana, Cuba
Let b√nc = m, then n ∈m2, m2+ 1, ..., m2+ 2m . Now we have two cases m is odd and m is even.
Assume m is odd and m ≥ 3, then
m odd ⇒ m | n ⇒ n ∈m2, m2+ m, m2+ 2m m odd ⇒ m − 2 | n. Now we have three sub-cases:
• m − 2 | m2 ⇒ m − 2 | (m − 2)2+ 4(m − 2) + 4 ⇒ m − 2 | 4 ⇒ m = 3.
• m − 2 | m2+ m ⇒ m − 2 | (m − 2)2+ 5(m − 2) + 6 ⇒ m − 2 | 6 ⇒ m = 3
or m = 5.
• m − 2 | m2+ 2m ⇒ m − 2 | (m − 2)2+ 6(m − 2) + 8 ⇒ m − 2 | 8 ⇒ m = 3.
From m = 3 we obtain n = 9, n = 12, n = 15.
From m = 5 we obtain n = 30 since 3 not divide 25 neither 35.
Assume m is even and m ≥ 6, then
m even ⇒ m − 1 | n ⇒ n ∈m2+ m − 2, m2+ 2m − 3 m even ⇒ m − 3 | n. Now we have two sub-cases:
• m−3 | m2+m−2 ⇒ m−3 | (m−3)2+7(m−3)+10 ⇒ m−3 | 10 ⇒ m = 8.
• m−3 | m2+2m−3 ⇒ m−3 | (m−3)2+8(m−3)+12 ⇒ m−3 | 12 ⇒ m = 6.
From m = 6 we obtain n = 45 since 3 not divide 40.
From m = 8 we not obtain n since 3 not divide 70 neither 77. Now just we need consider the trivial cases m = 1, m = 2, m = 4. From m = 1 we obtain n = 1, n = 2, n = 3.
From m = 2 we obtain n = 4, n = 5, n = 6, n = 7, n = 8. From m = 4 we obtain n = 18, n = 21, n = 24.
Finally, n ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 15, 18, 21, 24, 30, 45} Also solved by G.R.A.20 Math Problems Group, Roma, Italy.
J84. Al and Bo play the following game: there are 22 cards labeled 1 through 22. Al chooses one of them and places it on a table. Bo then places one of the remaining cards at the right of the one placed by Al such that the sum of the two numbers on the cards is a perfect square. Al then places one of the remaining cards such that the sum of the numbers on the last two cards played is a perfect square, and so on. The game ends when all the cards were played or no more card can be placed on the table. The winner is the one who played the last card. Does Al have a winning strategy?
Proposed by Titu Andreescu, University of Texas at Dallas, USA
First solution by Roberto Bosch Cabrera, University of Havana, Cuba
The winning strategy for Al is to choose the card labeled 2 in the first step. Note that the perfect squares in the play are: 4, 9, 16, 25, 36. We consider the equation 2 + m = n2 to obtain that Bo just can choose 7 or 14 in the second step.
If Bo chosses 7, then Al chosses 18. Bo cannot play since the equation 18 + m = n2 is not solvable, because 7 was chosen before. Thus Al is winner (2 − 7 − 18). If Bo chosses 14, then Al have another winning chain (2 − 14 − 11 − 5 − 20 − 16 − 9 − 7 − 18). At each step Bo have no choice other than choosing the card shown in the chain. So Al has a winning strategy and we are done.
Second solution by Ganesh Ajjanagadde, Acharya Vidya Kula, Mysore, India We claim that Al has a winning strategy.
On his first move, let Al choose 22. On his subsequent moves, let Al choose the maximum number available to him, such that the sum of his number and the previous number is a perfect square. It is clear that Bo has two choices for her first move, namely 3 and 14. Let us consider these two cases separately. Case 1. Bo chooses 14. Thus Bo has only one choice in each of her subsequent moves if Al sticks to his strategy. The sequence of moves are the following: 22, 14, 11, 5, 20, 16, 9, 7, 18. Once Al places 18, Bo has to either place 7, or 18, both of which are impossible.
Case 2. Bo chooses 3. The sequence of moves runs 22, 3, 13, 12, 4. Bo can now either place 5 or 21.
Case 2 a. Bo plays 5. Then the sequence of moves continues as follows: 5, 20, 16, 9, 7, 18. Once again we reach a state when Bo can make no further move.
Case 2 b Bo plays 21. Then the sequence of moves continues as follows: 21, 15, and Bo can now play either 1 or 10.
Case 2 b i Bo plays 1. The sequence continues thus: 1, 8, 17, 19, 6, 10. Now Bo can play either 15 or 6, both of which are not possible as they have been played earlier.
Case 2 b ii Bo plays 10. The sequence continues as follows: 10, 6, 19, 17, 8, 1. Now Bo has to play either 3, 8, or 15, none of which is possible as they have been played earlier. Thus Al can always force a win by sticking to this strategy.
Also solved by Daniel Campos Salas, Costa Rica; Daniel Lasaosa, Universidad Publica de Navarra, Spain; John T. Robinson, Yorktown Heights, NY, USA; Salem Malikic Sarajevo, Bosnia and Herzegovina.
Senior problems S79. Let an= 4 √ 2 +√n4, n = 2, 3, 4, . . . Prove that a1 5 + 1 a6 + 1 a12 + 1 a20 = 4 √ 8. Proposed by Titu Andreescu, University of Texas at Dallas
First solution by Simon Morris, Guernsey 1 a5 + 1 a6 + 1 a12 + 1 a20 = 1 214 + 2 2 5 + 1 214 + 2 2 6 + 1 214 + 2 2 12 + 1 214 + 2 2 20 = 1 21560 + 2 24 60 + 1 21560 + 2 20 60 + 1 21560 + 2 10 60 + 1 21560 + 2 6 60 = 1 21560(2 9 60 + 1) + 1 21560(2 5 60 + 1) + 1 21060(2 5 60 + 1) + 1 2606(2 9 60 + 1) = 2 −15 60 2609 + 1 + 2 −6 60 2609 + 1 + 2 −15 60 2605 + 1 + 2 −10 60 2605 + 1 = 2 −15 60 + 2− 6 60 2609 + 1 +2 −15 60 + 2− 10 60 2605 + 1 = 2 3 4(2−1+ 2− 51 60) 2(2−5160 + 2−1) +2 3 4(2−1+ 2− 55 60) 2(2−5560 + 2−1) = 2 3 4 2 + 234 2 = 2 3 4.
Second solution by Johnathon M. Ashcraft, Auburn Montgomery, USA Given the above information, we find
a5= 4 √ 2 +√5 4 = 214 + 2 2 5 = 2 5 20 + 2 8 20 = 2 1 4 1 + 2203 a6= 4 √ 2 +√64 = 214 + 2 1 3 = 2 15 60 + 2 20 60 = 2 1 4 1 + 2121 a12= 4 √ 2 + 12√ 4 = 214 + 2 1 6 = 2 3 12 + 2 2 12 = 2 1 4 1 + 2−121 a20= 4 √ 2 + 20√ 4 = 214 + 2 1 10 = 2 10 40 + 2 4 40 = 2 1 4 1 + 2−203 . Therefore
1 a5 + 1 a6 + 1 a12 + 1 a20 = = 1 214 1 + 2203 + 1 214 1 + 2121 + 1 214 1 + 2−121 + 1 214 1 + 2−203 = 1 214 1 1 + 2203 + 1 1 + 2−203 + 1 1 + 2121 + 1 1 + 2−121 = 1 214 2 + 2203 + 2− 3 20 2 + 2203 + 2− 3 20 +2 + 2 1 12 + 2− 1 12 2 + 2121 + 2− 1 12 ! = 2 214 = 234 = 4 √ 23 =√4 8.
Third solution by Prithwijit De, Kolkata, India We observe that a5= 2 1 4 + 2 2 5; a6 = 2 1 4 + 2 1 3; a12= 2 1 4 + 2 1 3; a20= 2 1 4 + 2 1 10.
Let u = 2601 = u and observe that
a5= u15+ u24= u15(u9+ 1); a6= u15+ u20= u15(u5+ 1); a12= u15+ u10= u10(u5+ 1); a20= u15+ u6 = u6(u9+ 1). Now, 1 a5 + 1 a6 + 1 a12 + 1 a20 = = 1 u15(u9+ 1) + 1 u15(u5+ 1)+ 1 u10(u5+ 1)+ 1 u6(u9+ 1) = 2 u15 = 4 √ 8, as desired.
Also solved by Arkady Alt, San Jose, California, USA; Brian Bradie, Christo-pher Newport University, USA; Daniel Lasaosa, Universidad Publica de Navarra, Spain; John T. Robinson, Yorktown Heights, NY, USA; Vicente Vicario Garcia, Huelva, Spain.
S80. Let ABC be a triangle and let Ma, Mb, Mc be the midpoints of sides BC, CA,
AB, respectively. Let the feet of the perpendiculars from vertices Mb, Mc in
triangle AMbMC be C2 and B1; the feet of the perpendiculars from vertices
Ma, Mb in triangle CMaMb be B2 and A1; the feet of the perpendiculars from
vertices Mc, Main triangle BMaMcbe A2and C1. Prove that the perpendicular
bisectors of B1C2, C1A2, and A1B2 are concurrent.
Proposed by Vinoth Nandakumar, Sydney University, Australia
Solution by Mihai Miculita, Oradea, Romania
Let A0, B0, C0be the midpoints of MbMc, MaMc, MaMband let a, b, c be the
per-pendicular bisectors of B1C1, C1A2, A1B2, respectively. Since triangle A0B0C0
is the complementary triangle of triangle MaMbMc and triangle MaMbMc is the
complementray triangle of ABC, triangles A0B0C0 and ABC are homothetic.
They have the same centroid G and at the same time G is the center of ho-mothety with the ratio 14. Let O and O0 be the centers of their circumcircles.
Thus
GO0=
1
4GO. (1)
Quadrilateral McMbB1C2 is cyclic, having center at A0, the midpoint of MbMc.
Thus A0B1 = A0C2, and triangle A0B1C2 is isosceles. This implies that A0∈ a.
On the other hand MbMckBC, thus OA⊥B1C2and since a⊥B1C2 we get OAka.
This means that line a in triangle A0B0C0 is homologous to the radius OA of
triangle ABC and so O0 ∈ a. Analoguously it can be proved that b and c pass
through O0, the circumcenter of triangle A0B0C0.
Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; Oleh Faynshteyn, Leipzig, Germany; Ricardo Barroso, University of Sevilla, Spain.
S81. Consider the polynomial P (x) = n X k=0 1 n + k + 1x k,
with n ≥ 1. Prove that the equation P (x2) = (P (x))2 has no real roots. Proposed by Dorin Andrica, Babes-Bolyai University, Romania
First solution by Daniel Campos Salas, Costa Rica Suppose there exist a real root t to the equation. Since
P (t2) ≥ 1
n + 1 > 0,
it follows that P (t2) = (P (t))2 > 0. From Cauchy-Schwarz we get
n X k=0 1 n + k + 1 ! n X k=0 1 n + k + 1t 2k ! ≥ n X k=0 1 n + k + 1t k !2 ,
which implies that
n X k=0 1 n + k + 1 ≥ 1. However, we have n X k=0 1 n + k + 1 < (n + 1) 1 n + 1 = 1,
a contradiction. It follows that the equation P (x2) = (P (x))2 has no real roots.
Second solution by John T. Robinson, Yorktown Heights, NY, USA
Consider the polynomial Q(x) = P (x2)−P (x)2. Since Q(0) = (n+1)1 −(n+1)1 2 > 0,
the problem is equivalent to proving the inequality P (x2) > P (x)2. Define the vector V (x) as
V (x) = xn √ 2n + 1, xn−1 √ 2n, . . . , 1 √ n + 1 .
Using vector dot product we have P (x2) = V (x) · V (x) = |V (x)|2 and P (x) = V (x) · V (1) = |V (x)| · |V (1)| · cos α,
where α is the angle between V (x) and V (1). Therefore P (x)2 = |V (x)|2· |V (1)|2cos2α.
Since cos2(a) ≤ 1, if we can show |V (1)|2 < 1 this will establish the inequality P (x2) > P (x)2. Note that |V (1)|2 = V (1) · V (1) = 1 2n + 1 + 1 2n + . . . + 1 n + 1.
It can be shown that |V (1)|2 < 1 for n ≥ 1 by induction on n. For n = 1, |V (1)|2 = 1/3 + 1/2 < 1. Next suppose that for some n:
|V (1)|2= 1 2n + 1+ 1 2n + . . . + 1 n + 1 < 1. Moving to n + 1 : |V (1)|2= 1 2n + 3 + 1 2n + 2 + . . . + 1 n + 2.
We see that we subtracted n+11 , and added 2n+31 and 2n+21 . But since n+11 >
1 2n+3+
1
2n+2, the value subtracted was larger than the values added, so |V (1)| 2
is decreasing for increasing n. Therefore P (x2) > P (x)2 for all n ≥ 1, and Q(x) has no real roots.
Third solution by Perfetti Paolo, Dipartimento di Matematica Tor Vergata Roma, Italy Since xk n + k + 1 = Z x 0 tk+n xn+1dt
the equation (P (x))2= P (x2) becomes 1 x2n+2 Z x2 0 tnt n+1− 1 t − 1 dt − 1 x2n+2 Z x 0 tnt n+1− 1 t − 1 dt 2 . = R(x) x2n+2
x = 0 is not a solution of the equation for any n by n+11 6= (n+1)1 2 for any n ≥ 1.
We show that the derivative of R(x)=. R0x2Q(t)dt − R0xQ(t)dt2 is positive for x > 0 and negative for x < 0. This is enough together with R(0) = 0.
R0(x) = 2xQ(x2) − 2Q(x) Z x 0 Q(t)dt = = 2x2n+1 n X k=0 x2k− 2n X q=0 xq q X r=0 1 n + 1 + r · 1 n + 2 + q − r =
= 2x2n+1 n X k=0 x2k 1 − q X r=0 1 n + 1 + r · 1 n + 2 + q − r − −x2n+1 n−1 X q=0 x2q+1 q X r=0 1 n + 1 + r · 1 n + 2 + q − r We observe that q X r=0 1 n + 1 + r 1 n + 2 + q − r ≤ n X r=0 1 n + 1 + r · 1 n + 2 + q − r ≤ n + 1 (n + 1)2 ≤ 1 2 (1) For x < 0 the conclusion R0(x) < 0 immediately follows.
As for x > 0, rewrite R0(x) as 2x2n+1 n X k=0 x2k1 2− q X r=0 1 n + 1 + r · 1 n + 2 + q − r + +2x2n+1 n X k=0 x2k 2 − 2x 2n+1 n−1 X q=0 x2q+1 q X r=0 1 n + 1 + r · 1 n + 2 + q − r. The first sum over k is positive by (1). Apart from the factor 2x2n+1we rewrite the second and the third sums as
n X k=0 x2k 4 + n X k=0 x2k 4 − n−1 X q=0 x2q+1 q X r=0 1 n + 1 + r · 1 n + 2 + q − r and by means of the AM-GM inequality we have
1 4x 2q+1 4x 2q+2≥ 1 2x 2q+1≥ x2q+1 q X r=0 1 n + 1 + r · 1 n + 2 + q − r which again in implied by (1). The proof is completed.
Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; Nguyen Manh Dung, Hanoi, Vietnam.
S82. Let a and b be positive real numbers with a ≥ 1. Further, let s1, s2, s3 be
nonnegative real numbers for which there is a real number x such that s1 ≥ x2, as2+ s3 ≥ 1 − bx.
What is the least possible value of s1+ s2+ s3in terms of a and b (the minimum
is taken over all possible values of x)?
Proposed by Zoran Sunic, Texas A&M University, USA
First solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain Clearly, s2+ s3 ≥ s2+ sa3 = as2a+s3 ≥ 1−bxa , where the first inequality reaches
equality iff a = 1 or s3 = 0. Therefore, s1+ s2+ s3 ≥ ax
2−bx+1
a ≥
4a−b2
4a2 , where
the second inequality reaches equality iff x = 2ab . We may consider two cases: (1) b2 ≤ 2a. Then, we may indeed take x = b
2a, s1 = x2, s2 = 2a−b2 2a2 ≥ 0 and s3= 0, yielding s1+ s2+ s3 = 4a−b 2 4a2 .
(2) b2 ≥ 2a. Take then x = 1 b, s1 =
1
b2, s2 = s3 = 0, for s1 + s2+ s3 = b12.
If a lower value were possible, then an x < 1b would exist such that 1−bxa ≤ s2+ s3 < b12− s1 ≤ b12− x2, or x would exist such that 0 > ab2x2− b3x + b2− a =
(bx − 1)(abx − b2+ a). Note that, if x < 1b, the first factor is negative, or the second factor must be positive, which is absurd, since abx−b2+a < ab1b−b2+a =
2a − b2 ≤ 0. Therefore, no value of s1+ s2+ s3 exists lower than the proposed
one in this case.
Finally note that, if b2 > 2a, then 1
b2 < 4a2a2 < 4a−b 2 4a2 , whereas if b2 < 2a, b12 > 2a 4a2 > 4a−b2
4a2 , and the least value that s1+ s2+ s3 may take is min
n 1 b2, 4a−b2 4a2 o , where x = min1b,2ab , s1 = min
n 1 b2, b2 4a2 o , s2 = max n 0,2a−b2a22 o , and s3 = 0.
Second solution by John T. Robinson, Yorktown Heights, NY, USA Since s1≥ x2 ≥ 0 we see that
√
s1 ≥ |x|. We may assume that x is nonnegative,
since we are minimizing s1+ s2+ s3 over all x, and for x ≥ 0, as2+ s3 ≥ 1 − bx
is a less restrictive constraint than as2+ s3 ≥ 1 + bx (which is the constraint
we would have if we replaced x with −x). Therefore 0 ≤ x ≤√s1, − x ≥ − √ s1, as2+ s3≥ 1 − bx ≥ 1 − b √ s1 and as2+ s3+ b √ s1 ≥ 1.
Since we are trying to minimize s1+ s2+ s3, and the LHS is the sum of terms
involving s1, s2, and s3 each of which increases for increasing s1, s2, or s3, the
minimum will be on the surface as2+ s3+ b
√
s1 = 1 in the octant s1, s2, s3 ≥ 0.
Note that since as2, s3, and b
√
s1 are all nonnegative, we require as2, s3, and
b√s1 ≤ 1. Next consider the s2+ s3 component of s1+ s2+ s3 - for any given
value of s1, we have as2+ s3= 1 − b
√
s1, and since a ≥ 1, s2+ s3is minimized by
choosing s3= 0. So the problem is now simplified to minimizing s1+ s2 subject
to the constraint as2+ b
√ s1 = 1.
Since √s1= (1 − as2)/b, we have s1+ s2= (1−as2)
2
b2 + s2. Taking the derivative
of the RHS, we find the minimum when −2a(1−as2)
b2 + 1 = 0, b 2 2a = 1 − as2, s2 = 1a − b 2 2a2. So if 1a ≥ b 2 2a2, or 2a ≥ b2, then s1 + s2 + s3 is minimized by choosing s2 = 1 a− b2 2a2, s1 = 1 − as2 b 2 = b 2 4a2, s3 = 0.
Otherwise, if b2 > 2a, the minimum of the quadratic (1−as2)2
b2 + s2 occurs for
a negative value of s2, and is at its minimum value for nonnegative s2 when
s2= 0. Therefore s1+ s2+ s3 is minimized by choosing
s1 =
1
S83. Find all complex numbers x, y, z of modulus 1, satisfying y2+ z2 x + x2+ z2 y + x2+ y2 z = 2(x + y + z).
Proposed by Cosmin Pohoata, Bucharest, Romania
First solution by Magkos Athanasios, Kozani, Greece ewrite the given relation as
(x2+ y2+ z2) 1 x+ 1 y + 1 z = 3(x + y + z).
Since the numbers are of unit modulus we obtain (x2+ y2+ z2)(¯x + ¯y + ¯z) = 3(x+y+z). Passing to the moduli we get either |x2+y2+z2| = 3 or x+y+z = 0. If x + y + z = 0, because |x| = |y| = |z| = 1, the images of the numbers are the vertices of an equilateral triangle inscribed in the unit circle. Set x = cos a + i sin a, a ∈ [0, 2π). Then
y = x(cos 120◦+ i sin 120◦) = x −1 2 + i √ 3 2 ! , z = x(cos 240◦+ i sin 240◦) = x −1 2 + i −√3 2 ! . If |x2+ y2+ z2| = 3, we have |x2+ y2+ z2| = |x2| + |y2| + |z2|.
In this case, it is known that there exist positive reals A, B such that x2 = Ay2
and x2 = Bz2. Since |x| = |y| = |z| = 1 we get A = B = 1, hence, x2= y2 = z2. This means that we have one of the following possibilities:
x = y = z, x = y = −z, x = −y = z, x = −y = −z,
where |x| = 1. It is easy to verify that all of of the above triplets satisfy the initial condition.
Second solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain Clearly, x +y2+zx 2 = x2+yx2+z2 = x x2+ y2+ z2, where x denotes the complex conjugate of x, and we have used that the modulus of x is 1 = xx. Therefore, the given equation is equivalent to
Taking the complex conjugate of both sides of this last equation and using it for simplification yields
9 (x + y + z) = 3 (x + y + z) x2+ y2+ z2 = (x + y + z) x2+ y2+ z2
2
. The given equation may have a solution then in one of the following two cases: 1) x2+ y2+ z2 = 3 = x2 + y2 + z2
. Now, equality in the triangular inequality forces x2, y2, z2 to define collinear vectors in the complex plane, or x2 = y2 = z2. Note that any such triplet yields in fact a solution to the given equation, since y2+zx 2 = 2xx2 = 2x, and similarly for the other two fractions. 2) x + y + z = 0. Then, y2+zx 2 = (y+z)
2−2yz
x = x −
2yz
x , and similarly for the
other two fractions. Substitution in the original equation yields 0 = yz x + zx y + xy z = xyz x2+ y2+ z2,
and since xyz 6= 0, then x2+y2+z2= 0. Now, x2 = −y2−z2= −(y+z)2+2yz =
−x2+ 2yz, and x2 = yz, and similarly y2 = zx, z2 = xy. Substitution in the
given equation yields y2+zx 2 = z + y, and similarly for the other two fractions, or all triplets x, y, z such that x + y + z = 0 and x2+ y2+ z2 = 0 are solutions of the given equation. Now, xy + yz + zx = (x+y+z)2−x2 2−y2−z2 = 0, and thus by Cardano-Vieta relations, x, y, z are the three roots of an equation of the form r3− a = 0, where a = xyz is a complex number with modulus 1, ie, x, y, z are the three cubic roots of a complex number of modulus 1.
Therefore, all the solutions of the given equation either satisfy x2= y2 = z2 (ie, x = ±y and x = ±z), or x, y, z are the three cubic roots of any complex number of modulus 1.
Third solution by Perfetti Paolo, Dipartimento di matematica Universita degli studi di Tor Vergata, Italy
Clearing the denominators we obtain
(xy + yz + zx)(x2+ y2+ z2) − 3(xyz)(x + y + z) xyx = 0 and then 1 x + 1 y + 1 z (x2+ y2+ z2) = 3(x + y + z)
Taking the absolute value and observing that for complex numbers of modulus one the following holds
|x + y + z| = 1 x + 1 y + 1 z ,
we get |x2+ y2+ z2| = 3 or x + y + z = 0 and this is possible only when the three complex numbers are collinear or they are vertices of a equilateral triangle inscribed in the unit circle.
S84. Let ABC be an acute triangle and let ω and Ω be its incircle and circumcircle, respectively. Circle ωA is tangent internally to Ω at A and tangent externally
to ω. Circle ΩA is tangent internally to Ω at A and tangent internally to ω.
Denote by PAand QAthe centers of ωAand ΩA, respectively. Define the points
PB, QB, PC, QC analogously. Prove that
PAQA BC + PBQB CA + PCQC AB ≥ √ 3 2 .
Proposed by Cezar Lupu, Univeristy of Bucharest, Romania
First solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain Denote by R, r, RA, rA the respective radii of Ω, ω, ΩA, ωA. It is clear by
con-struction that PA, QA are on segment OA, and APA = rA, AQA = RA,
yielding PAQA = RA − rA. Denote by A0, A00 the respective points where
ωA, ΩA touch ω, it is clear that A0, A00 are respectively on lines IPA, IQA, and
IPA = IA0 + A0PA = r + rA, IQA = QAA00− IA00 = RA− r. Furthermore,
∠IAPA= ∠IAQA= ∠IAO = ∠IAC − ∠OAC = A2−π2+ B = B−C2 . Therefore,
using the theorem of the cosine,
r2+ rA2 + 2rrA= IPA2 = IA2+ APA2− 2IA · APAcos ∠IAPA
= r 2 sin2 A2 + r 2 A− 2 rrAcosB−C2 sinA2 , r cos2A 2 = 2rAsin A 2 sinA 2 + cos B − C 2 = rrA R cosB2 cosC2 sinB2 sinC2 ; r2+ R2A− 2rRA= IQA2 = IA2+ AQ2A− 2IA · AQAcos ∠IAQA
= r 2 sin2 A2 + R 2 A− 2 rRAcosB−C2 sinA2 , r cos2 A 2 = 2RAsin A 2 cosB − C 2 − sin A 2 = rRA R . We have used that sinA2 = cosB+C2 and r = 4R sinA2 sinB2 sinC2. Then,
PAQA= RA− rA= R cos2 A 2 1 − sinB2 sinC2 cosB2 cosC2 ! = R sin A 2 cos 2 A 2 cosB2 cosC2 ; PAQA BC = cosA2 4 cosB2 cosC2 = tanB2 + tanC2 4 ,
and similarly PBQB CA = tanC2+tanA2 4 and PCQC AB = tanA2+tanB2 4 .
It suffices therefore to prove that tanA2 + tanB2 + tanC2 ≥ √3. But it is well known (or easily provable using that A + B + C = π) that
tanA 2 tan B 2 + tan B 2 tan C 2 + tan C 2 tan A 2 = 1,
while application of the inequality between arithmetic and quadratic means for any positive real numbers u, v, w yields
(u + v + w)2 ≥ (u + v + w)
2
3 + 2(uv + vw + wu),
with equality iff u = v = w, or u + v + w ≥ p3(uv + vw + wu). The result follows, with equality holding if and only if A = B = C, i.e. if and only if triangle ABC is equilateral.
Undergraduate problems
U79. Let a1= 1 and an= an−1+ ln n. Prove that the sequencePni=1a1i is divergent.
Proposed by Ivan Borsenco, University of Texas at Dallas, USA
First solution by Arin Chaudhuri, North Carolina State University, NC, USA Note an= a1+ (a2− a1) + · · · + (an− an−1) = 1 + ln(2) + · · · + ln(n) = 1 + ln(n!).
From Stirling’s approximation we have ln n! = ln( √ 2π) − n + n ln n + 1 2ln n + cn where cn→ 0. Hence, an= C − n + n ln n + 1 2ln n + cn where C = 1 + ln(√2π).
If n ≥ 2 then diving throughout by n ln n we have an n ln n = C n ln n − 1 ln n+ 1 + 1 2n+ cn n ln n Note all terms above vanish as n → ∞ except 1. Hence
lim
n→∞
an
n ln n = 1 (1)
Hence we can find an N such that for all n ≥ N an
n ln n ≤ 2 Hence for all n ≥ N
1 2n ln n ≤
1 an
Using the well known result thatP∞
k=2k ln k1 = +∞, we have
P∞
k=1a1k = +∞.
Second solution by Arkady Alt, San Jose ,California, USA Since n! < n2n and an− a1 = n P k=2 (ak− ak−1) = n P k=2 ln k = ln n! we get an= 1 + ln n!.
Note that an< 1 + n ln n 2 < n ln n, n ≥ 2 ⇐⇒ 1 an > 1 n ln n, n ≥ 2. Moreover, 1 an
> ln ln(n + 1) − ln ln n, for n ≥ 2 because by the Mean Value Theorem for some cn∈ (n, n + 1) we have
ln ln(n + 1) − ln ln n = 1 cnln cn < 1 n ln n. Hence, n X k=1 1 an = 1 + n X k=2 1 an > 1 + n X i=2 (ln ln(k + 1) − ln ln k) = = 1 + ln ln(n + 1) − ln ln 2, and, therefore, sequence
n P k=1 1 an is divergent.
Third solution by Jean Mathieux, Senegal
We have that a3 ≤ 3 ln 3. Suppose that for n > 3, an≤ n ln n, then
an+1≤ n ln n + ln(n + 1) ≤ (n + 1) ln(n + 1). So for all i > 2,a1 i ≥ 1 i ln 1. Also, since t → 1 t ln t is decreasing, Ri+1 i 1 t ln tdt ≤ 1 i ln i. Thus n X i=3 1 ai ≥ Z n+1 3 1 t ln tdt = ln(ln(n + 1)) − ln(ln(3)). Hence the given sequence is divergent.
Also solved by Magkos Athanasios, Kozani, Greece; Brian Bradie, Christopher Newport University, USA; Daniel Lasaosa, Universidad Publica de Navarra, Spain; John T. Robinson, Yorktown Heights, NY, USA; Perfetti Paolo, Di-partimento di matematica Universita degli studi di Tor Vergata, Italy; Vicente Vicario Garcia, Huelva, Spain; Roberto Bosch Cabrera, Faculty of Mathemat-ics, University of Havana, Cuba.
U80. Let f : R → R be a differentiable function at the origin satisfying f (0) = 0 and f0(0) = 1. Evaluate lim x→0 1 x ∞ X n=1 (−1)nfx n .
Proposed by Titu Andreescu, University of Texas at Dallas, USA
First solution by Brian Bradie, Christopher Newport University, USA Because f (0) = 0 and f0(0) = 1, we have
lim x→0 1 xf x n = lim x→0 f (x/n) − f (0) x = lim w→0 f (w) − f (0) nw = 1 nf 0(0) = 1 n. Thus, lim x→0 1 x ∞ X n=1 (−1)nfx n = ∞ X n=1 (−1)nlim x→0 1 xf x n = ∞ X n=1 (−1)n1 n = − ∞ X n=1 (−1)n−11 n = − ln 2.
Second solution by John T. Robinson, Yorktown Heights, NY, USA
By definition, the limit as x goes to 0 of (f (x) − f (0))/x = f (x)/x is f0(0) = 1. Consider f (x/n)/x: substituting y = x/n, f (x/n) x = f (y) n · y = 1 n· f (y) y ,
therefore the limit as x goes to 0 of f (x/n)x = n1 · f0(0) = 1n. It follows that the limit being asked for is
−1 + 1/2 − 1/3 + 1/4 − 1/5 + · · · = − ln 2.
Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; Perfetti Paolo, Dipartimento di matematica Universita degli studi di Tor Vergata, Italy.
U81. The sequence (xn)n≥1 is defined by
x1 < 0, xn+1 = exn− 1, n ≥ 1.
Prove that limn→∞nxn= −2.
Proposed by Dorin Andrica, Babes-Bolyai University, Romania
First solution by Ovidiu Furdui, The University of Toledo, OH
By induction it can be proved that xn < 0 for all n ≥ 1. On the other hand,
since ex− 1 ≥ x for all x ∈ R, we get that xn+1 = exn− 1 ≥ xn, and hence,
the sequence increases. It follows that x1 < xn < 0 for all n ≥ 1, and hence,
the sequence converges. If l = lim xn, then passing to the limit as n → ∞ in
the recurrence relation we obtain that l = el − 1 from which it follows that l = 0. We calculate lim
n→∞nxn by using Cesaro-Stolz lemma. We have, since
lim x→0 ex−1 x = 1 and limx→0 x2 ex−1−x = 2, that lim n→∞nxn= − limn→∞ n −1 xn = − lim n→∞ 1 1 xn − 1 xn+1 = − lim n→∞ xn+1· xn xn+1− xn = − lim n→∞ exn− 1 xn · lim n→∞ x2n exn− 1 − x n = −2, and the problem is solved.
Second solution by Brian Bradie, Christopher Newport University, USA If xn < 0, then xn+1 = exn − 1 < 1 − 1 = 0. As we are given that x1 < 0,
it follows by induction on n that xn < 0 for all n. Moreover, it is clear that
limn→inf tyxn= 0 for any x1< 0. Now, let yn= −12xn. Then, yn> 0 for all n,
yn→ 0 and yn+1= − 1 2 e −2yn− 1 = y n− y2n+ O(y3n).
According to formula (8.5.3) (N. G. de Bruijn, Asymptotic Methods in Analysis, Dover Publications Inc., New York, 1981, page 155) it follows that
yn= 1 n+ O(n −2ln n) as n → ∞. Therefore, xn= − 2 n+ O(n −2 ln n) as n → ∞, and limn→∞nxn= −2.
Solution to Mathematical Reflections Problem U81 Brian Bradie Department of Mathematics Christopher Newport University Newport News, VA
Solution to Mathematical Reflections Problem U81
Problem: The sequence (xn)n≥1is defined by
x1< 0, xn+1= exn− 1, n ≥ 1.
Prove that limn→∞nxn= −2.
Solution: If xn < 0, then xn+1 = exn− 1 < 1 − 1 = 0. As we are given that
x1< 0, it follows by induction on n that xn< 0 for all n. Moreover, the cobweb
diagram shown below illustrates that xn→ 0 for any x1< 0.
x1 x2 x3 x4
x5
xn+1 = xn
xn+1 = exn – 1
Now, let yn= −12xn. Then, yn> 0 for all n, yn→ 0 and
yn+1= −
1 2 e
−2yn− 1 = y
n− yn2+ O(yn3).
According to formula (8.5.3) in [1, page 155], it follows that yn= 1 n+ O(n −2 ln n) 4
Also solved by Arin Chaudhuri, North Carolina State University, NC, USA; Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Universidad Publica de Navarra, Spain; G.R.A.20 Math Problems Group, Roma, Italy; John T. Robin-son, Yorktown Heights, NY, USA; Perfetti Paolo, Dipartimento di matematica Universita degli studi di Tor Vergata, Italy.
U82. Evaluate lim n→∞ n Y k=1 1 +k n n k3 .
Proposed by Cezar Lupu, Univeristy of Bucharest, Romania
First solution by Brian Bradie, Christopher Newport University, USA Let y = n Y k=1 1 + k n n/k3 . Then, ln y = n X k=1 n k3 ln 1 + k n = n X k=1 n k3 ∞ X `=1 (−1)`−1(k/n) ` ` ! = ∞ X `=1 (−1)`−1n 1−` ` n X k=1 k`−3 ! = n X k=1 1 k2 − 1 2n n X k=1 1 k+ ∞ X `=3 (−1)`−1n 1−` ` n X k=1 k`−3 ! . Now, as n → ∞, Pn k=1k`−3= O(n`−2), so ∞ X `=3 (−1)`−1n 1−` ` n X k=1 k`−3 ! = O(n−1) ∞ X `=3 (−1)`−1 ` → 0. Additionally, as n → ∞, Pn k=1k1 = O(ln n), so 1 2n n X k=1 1 k = O ln n n → 0. Finally, as n → ∞, n X k=1 1 k2 → ∞ X k=1 1 k2 = π2 6 . Thus, lim n→∞ln y = π2 6 and n→∞lim n Y k=1 1 +k n n/k3 = eπ2/6.
Second solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain Call Pn=Qnk=1 1 +kn n k3. Then, ln Pn= n X k=1 n k3 ln 1 + k n .
But since ln (1 + x) = x −x22 +x33 −x44 + ..., we may write n k3 ln 1 +k n = 1 k2 − 1 2nk+ 1 3n2 − ...
The terms in the sum in the RHS have alternating signs and decreasing absolute value, or 1 k2 − 1 2nk + 1 3n2 > n k3 ln 1 + k n > 1 k2 − 1 2nk. Adding over k, 1 3n > ln Pn− n X k=1 1 k2 + 1 2n n X k=1 1 k > 0.
As n tends to infinity, the upper bound tends to 0, or the middle term has limit 0. Now, ln(k) − ln(k − 1) = Z k k−1 1 xdx > 1 k > Z k+1 k 1 xdx = ln(k + 1) − ln(k), and 1+ln(n)2n > 2n1 Pn k=11k > ln(n+1)
2n , and since the upper and lower bounds
tend to 0 as n grows, we conclude that limn→∞Pn=P∞k=1k12, which we may
recognize as ζ(2) = π62. It follows that lim n→∞ n Y k=1 1 + k n n k3 = eζ(2)= eπ26 .
Third solution by John Mangual, New York, USA The limit is eπ26 . Let S denote the limit:
S = lim n→∞ n Y k=1 1 +k n n k3 (2) Instead of evluating S directly, let’s examine the logarithm. By continuity of the logarithm we can write
ln S = lim n→∞ n X k=1 n k3ln 1 +k n (3)
Huristically speaking each term behaves like 1/k2therefore we will subtract our guess from the summation:
ln S −π 2 6 = limn→∞ n X k=1 1 k2 n kln 1 + k n − 1 (4)
By Taylor’s theorem we can estimate the logarithm. For any x ∈ [0, 1] there exists ξ ∈ [0, x] such that:
| ln(1 + x) − (x + x2/2)| < x
3
(1 + ξ)2 ≤ x
3 (5)
This is a uniform bound on the difference. Therefore we can rewrite (3):
n X k=1 1 k2 n k k n− k2 n2 − 1 + n X k=1 1 k2 n k ln 1 + k n − k n− k2 n2 (6)
By triangle inequality the second term is bounded by:
n X k=1 1 k2 k n 2 = 1 n (7)
The second term also converges, despite the harmonic series:
n X k=1 1 k2 n k k n− k2 n2 − 1 = 1 n n X k=1 1 k = o ln n n (8)
Both (6) and (7) are bounded as n → ∞ so ln S − π2/6 = 0.
Also solved by Perfetti Paolo, Dipartimento di matematica Universita degli studi di Tor Vergata, Italy; Arin Chaudhuri, North Carolina State University, Raleigh, NC; Arkady Alt , San Jose ,California, USA; G.R.A.20 Math Prob-lems Group, Roma, Italy; John T. Robinson, Yorktown Heights, NY, USA; Vicente Vicario Garcia, Huelva, Spain.
U83. Find all functions f : [0, 2] → (0, 1] that are differentiable at the origin and satisfy f (2x) = 2f2(x) − 1, for all x ∈ [0, 1].
Proposed by Titu Andreescu, University of Texas at Dallas, USA
First solution by Li Zhou, Polk Community College, USA Let g(x) = arccos f (x) for all x ∈ [0, 2]. Then for x ∈ [0, 1],
cos g(2x) = f (2x) = 2 cos2g(x) − 1 = cos(2g(x)),
thus g(2x) = 2g(x). Hence, for any x ∈ [0, 2], g(x) = 2g(x/2) = 4g(x/4) = · · · = 2ng(x/2n) for all n ≥ 1. Since g is differentiable at 0, lim
n→∞ g(x/2
n)
x/2n = k
for some constant k. Therefore, g(x) = kx for all x ∈ [0, 2]. Considering the range of f , we conclude that f (x) = cos(kx) with −π/4 < k < π/4. Finally, it is easy to verify that all such functions f do satisfy the conditions.
Second solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain Taking x = 0 yields f (0) = 2f2(0) − 1, or f (0) is a root of 0 = 2r2− r − 1 =
(2r + 1)(r − 1). Since f (0) ∈ (0, 1], then f (0) = 1. Take now any > 0. Obviously, f (2) − f (0) 2 = f2() − f2(0) = (f () + f (0)) f () − f (0) .
If f is differentiable at the origin, it is obviously continuous at the origin, and when → 0, the previous equality becomes f0(0) = 2f (0)f0(0) = 2f0(0), or f0(0) = 0. This relation is obviously equivalent to limx→0f (x)x = 0.
The interval (0, 2] may be defined as the disjoint union of sets of the form Ay = y,y2,y4,y8, ... , where y takes on all possible real values in (1, 2]. Note
that this is true since (1) Ay∩ Ay0 = ∅ iff y 6= y0, since if an element belongs to
both Ay and Ay0, then non-positive integers a, a0exist such that x = 2ay = 2a 0
y0, and since 2y > 2 ≥ y0, and 2y0 > 2 ≥ y, then a = a0 and y = y0, and (2) for any x ∈ (0, 2], sufficiently high integral exponents a yield 2ax > 1. The minimum of all such exponents will obviously yield 2ax = y for some y ∈ (1, 2], and x ∈ Ay.
Take now any function (not necessarily continuous) g : (1, 2] → (0, 1]. We will now construct a function f : [0, 2] → (0, 1] such that f (0) = 1, which satisfies the given functional equation, and such that
(1) if x ∈ (1, 2], then f (x) = g(x), (2) limx→0f (x) = 1 and
The last two conditions are equivalent to f being differentiable at the origin with value 1 and derivative 0. In order to construct f , for any y ∈ (1, 2], take f (y) = g(y); we may write f (y) = cos(αyy) for some αy ∈ 0,π2 since
f (y) ∈ (0, 1]. Now, fy 2 = r 1 + cos(αyy) 2 = r cos2αyy 2 = cosαy y 2 ,
where we have selected the positive root since f (x) ∈ (0, 1] for all x, and a trivial exercise in induction yields f (x) = cos(αyx) for all x ∈ Ay. Repeat the
same procedure for all y ∈ (1, 2]. It is obvious that f thus constructed satisfies the given functional equation, and condition (1). Conditions (2) and (3) are also easily checked, since, as it is well known, cos(β) → 1 and cos(β)−1β → 0 for any β → 0. Therefore, for any g(x) defined over (1, 2], a function f has been constructed that satisfies the requirements of the problem, and given any function f that satisfies the requirements of the problem, its values in (1, 2] biunivocally determine all values of f in (0, 1]. Therefore, for any such g an f may be constructed, and no other solutions may exist.
U84. Let f be a three times differentiable function on an interval I, and let a, b, c ∈ I. Prove that there exists ξ ∈ I such that
f a + 2b 3 +f b + 2c 3 +f c + 2a 3 −f 2a + b 3 −f 2b + c 3 −f 2c + a 3 = = 1 27(a − b)(b − c)(c − a)f 000(ξ).
Proposed by Vasile Cirtoaje, University of Ploiesti, Romania
First solution by Arkady Alt, San Jose, California, USA Let g (t) := f a + b + c 3 + t − f a + b + c 3 − t and let x = b − c 3 , y = c − a 3 , z = a − b
3 then x + y + z = 0 and δ (a, b, c) := f
a + 2b 3 + f b + 2c 3 + f c + 2a 3 − f 2a + b 3 − f 2b + c 3 − f 2c + a 3 = g (x) + g (y) + g (z) . We will consider non-trivial case where x, y, z 6= 0.
Note that if I = (p, q) then x, y, z ∈ (p1, q1) where p1:= p −
a + b + c
3 and
q1 := q −
a + b + c
3 and g is three times differentiable function on the inter-val (p1, q1) .
Since g (0) = 0 and g00(0) = 0 then by Maclaurin’s Theorem (1) g (t) = g0(0) t + g 000(θ) t3 6 fore some θ ∈ (p1, q1) . Applying (1) to t = x, y, z we obtain g (x)+g (y)+g (z) = g 000(θ x) x3+ g000(θz) y3+ g000(θz) z3 6 ( because x+y+z = 0). Since g000(t) := f000 a + b + c 3 + t + f000 a + b + c 3 − t then δ (a, b, c) = 1 6 P cyc x3 f000 a + b + c 3 + x + f000 a + b + c 3 − x and by Darboux’s Theorem about intermediate values of derivative for differentiable function f00 we there is such ξ ∈ I such that
P cyc x3 f000 a + b + c 3 + x + f000 a + b + c 3 − x = 2 x3+ y3+ z3 f000(ξ) . Thus δ (a, b, c) = x 3+ y3+ z3 f000(ξ) 3 and, because x 3+ y3+ z3 = 3xyz, we finally obtain δ (a, b, c) = 1 27(a − b) (b − c) (c − a) f 000(ξ) .
Second solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain Note first of all that we may choose wlog c > b > a, since exchanging any two of these values, inverts the sign of both sides of the given equation. Define now, for m 6= 0, and suitable parameters to be defined later ∆1, ∆2 > 0, functions
f3(x), g3(x): f3(x) = f (x + ∆1+ ∆2) − f (x − ∆1+ ∆2) − f (x + ∆1− ∆2) + f (x − ∆1− ∆2) 4∆1∆2 , g3(x) = m(x − x3) + h3.
Assume now that x3 and ∆3> 0 are chosen in a way such that
H3 = (x3− ∆3, x3+ ∆3) ⊂ I.
Obviously, f3(x) and g3(x) are differentiable in the interval H3. Therefore, by
Cauchy’s generalization of the mean value theorem, x2∈ H3 exists such that
f30(x2) = f3(x3+ ∆3) − f3(x3− ∆3) g3(x3+ ∆3) − g3(x3− ∆3) g30(x2) = f3(x3+ ∆3) − f3(x3− ∆3) 2∆3 . Using now this value of x2, define functions f2(x), g2(x):
f2(x) = f0(x + ∆1) − f0(x − ∆1) 2∆1 , g2(x) = m(x − x2) + h2. Note that f30(x) = f2(x + ∆2) − f2(x − ∆2) 2∆2
Assume again that ∆2 is chosen such that
H2 = (x2− ∆2, x2+ ∆2) ⊂ I.
Again, f2(x) and g2(x) are differentiable in H2, and x1 ∈ H2 exists such that
f20(x1) = f2(x2+ ∆2) − f2(x2− ∆2) g2(x2+ ∆2) − g2(x2− ∆2) g02(x1) = f2(x2+ ∆2) − f2(x2− ∆2) 2∆2 =
= f30(x2).
Using now this value of x1, define finally functions f1(x) = f00(x) and g1(x) =
m(x − x1) + h1. Note that
f20(x) = f1(x + ∆1) − f1(x − ∆1) 2∆1
. If once more ∆1 is chosen so that
H1 = (x1− ∆1, x1+ ∆1) ⊂ I,
then f1(x) and g1(x) are differentiable in H1, and ξ ∈ H1 exists such that
f10(ξ) = f1(x1+ ∆1) − f1(x1+ ∆1) g1(x1+ ∆1) − g1(x1− ∆1) g01(ξ) = f1(x1+ ∆1) − f1(x1− ∆1) 2∆1 = = f20(x1).
Therefore, we have proved that ξ ∈ H1 ⊂ I exists such that
f000(ξ) = f10(ξ) = f20(x1) = f30(x2) =
f3(x3+ ∆3) − f3(x3− ∆3)
2∆3
,
for suitably defined x3, ∆1, ∆2, ∆3. Taking ∆1 = c−b6 , ∆2 = b−a6 , ∆3 = c−a6 ,
x3= a+b+c3 , it follows that
H1, H2, H3 ⊂ 2a + b 3 , 2c + b 3 ⊂ I,
and inserting these very values into the form of f3(x) substituted in the
expres-sion for f000(ξ), the conclusion follows. Note that this general process may be used to find other possible values of f000(x) in I by selecting other values for x3, ∆1, ∆2, ∆3 (always, of course, values such that H1, H2, H3 ∈ I), and that
for functions differentiable more than three times, the process may be carried on in the same way.
Olympiad problems
O79. Let a1, a2, . . . , anbe integer numbers, not all zero, such that a1+a2+. . .+an= 0.
Prove that
|a1+ 2a2+ . . . + 2k−1ak| >
2k 3, for some k ∈ {1, 2, . . . n}.
Proposed by Bogdan Enescu, “B.P.Hasdeu” National College, Romania
First solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain Assume that a1+ 2a2+ ... + 2k−1ak
≤ 2
k
3 for all k ∈ {1, 2, ..., n}, the ai being
integers. We shall prove by induction that a1 = a2 = ... = an = 0. For k = 1,
the result is trivial, since |a1| ≤ 23 < 1 directly results in a1 = 0. If the result is
true for i = 1, 2, ..., k − 1, then 2k 3 ≥ a1+ 2a2+ ... + 2 k−1a k = 2 k−1|a k|,
yielding |ak| ≤ 23 < 1, and again ak= 0. All the ai are then zero, which is not
true. The result follows.
Second solution by John T. Robinson, Yorktown Heights, NY, USA
Let k be the smallest integer such that ak is non-zero, that is, a1 = a2= . . . =
ak−1 = 0 (if k > 1) and |ak| > 0. Since ak is an integer, |ak| ≥ 1. Therefore
|a1+ 2a2+ . . . + 2k−1ak| = 2k−1|ak|, and 2k−1|ak| ≥ 2k−1> 23 · 2k−1 = 2
k
O80. Let n be an integer greater than 1. Find the least number of rooks such that no matter how they are placed on an n × n chessboard there are two rooks that do not attack each other, but at the same time they are under attack by third rook.
Proposed by Samin Riasat, Notre Dame College, Dhaka, Bangladesh
First solution by Kee-Wai Lau, Hong Kong, China
We show that the least number of rooks is 2n − 1. The standard algebraic notation of the n × n chessboard is used. By placing the rooks on a12, a13, . . .,
a1n, a21, a31, . . ., an1, we see that 2n − 2 rooks are not sufficient. We will prove
by induction that 2n − 1 rooks are sufficient. For n = 2, the result is clear. We now suppose that the result is true for n = k ≥ 1. By placing the 2k + 1 rooks on the (k + 1) × (k + 1) chessboard, there is at least one row containing one rook or no rooks. Otherwise the total number of rooks is greater than or equal to 2k + 2, which is not true. Similarly there is at least one column containing one rook or no rooks. Select any such row and any such column and delete them from the (k + 1) × (k + 1) chessboard. We combine the undeleted parts of the (k + 1) × (k + 1) chessboard to obtain a k × k chessboard which contains at least 2k − 1 rooks. Select any 2k − 1 rooks. By the induction assumption, they are sufficient. It follows that 2k − 1 rooks are sufficient for the (k + 1) × (k + 1) chessboard. This completes the solution.
Second solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain We will show by induction that 2n − 1 rooks are enough to guarantee that one rook attacks two rooks that do not attack each other. For n = 2, the result is obvious, since 3 rooks can only be placed in such a way that one row contains 2 rooks and one column contains 2 rooks, and the rook in the intersection of this row and this column attacks the other two, which do not attack each other. Assume the result is true for n = 2, 3, ..., m, and consider 2m + 1 rooks in an (m + 1) × (m + 1) board. Clearly, there is at least one row with at most one rook (otherwise there would be at least 2m + 2 rooks), and analogously at least one column with at most one rook. Eliminate one such row and one such column, and the rooks contained therein, resulting in an m × m board with at least 2m + 1 − 2 = 2m − 1 rooks. Now, if two rooks attacked each other before the elimination, they were either in the same row or in the same column, so if the survived the elimination, they still attack each other. Conversely, if they attack each other after the elimination, they also attacked each other before the elimination. But by hypothesis of induction, there is after the elimination one rook that attacks two which do not attack each other, and the situation was exactly the same before the elimination, completing the proof.
Let us now see that 2n − 2 rooks are not enough to guarantee that one rook attacks two rooks which do not attack each other: consider all squares in the first row and first column filled, except for the intersection of the first row and first column. Obviously, two rooks attack each other iff either they are both in the first row, or they are both in the first column, and if one rook attacks two, then these two also attack each other, since the intersection of the first row and the first column is empty in the proposed distribution. So 2n − 2 rooks do not guarantee the desired arrangement, and 2n − 1 is the number that we are looking for.
Third solution by G.R.A.20 Math Problems Group, Roma, Italy
We will show that the least number of rooks such that the property holds in a m × n chessboard is m + n − 1.
If the rooks are less than m + n − 1, we can place them along the first column and along the first row but not at the top left corner (there are m + n − 2 places). The property does not hold for this displacement.
The thesis holds trivially when m + n ≤ 6, n > 1 and m > 1. Now we consider a m × n chessboard with m + n > 6, n > 1 and m > 1. and we assume that our thesis holds for any m0× n0 chessboard such that m0+ n0 < m + n and m0 > 1,
n0 > 1. We can assume without loss of generality that n ≥ m and therefore n > 3. Since we have at least m + n − 1 ≥ m + 1 rooks, then there is a row with at least two rooks. If there is at least another rook in the corresponding colums then the property holds. Otherwise we can cancel these two columns obtaining a m × (n − 2) chessboard with at least m + (n − 2) − 1 rooks. Since m + n > m + (n − 2), m > 1 and n − 2 > 1, by the inductive hypothesis, the property holds in this smaller checkboard and therefore it holds also in the initial one.
O81. Let a, b, c, x, y, z ≥ 0. Prove that
(a2+ x2)(b2+ y2)(c2+ z2) ≥ (ayz + bzx + cxy − xyz)2.
Proposed by Titu Andreescu, University of Texas at Dallas
First solution by Daniel Campos Salas, Costa Rica
The inequality is trivial if any of x, y, z equals 0. Suppose that xyz 6= 0. Therefore, dividing by (xyz)2 it follows that the inequality is equivalent to
(m2+ 1)(n2+ 1)(p2+ 1) ≥ (m + n + p − 1)2,
where (m, n, p) = xa,yb,cz. After expanding and rearranging some terms it follows that this inequality is equivalent to
m2n2p2+ (m2n2+ m + n) + (n2p2+ n + p) + (p2m2+ p + m) ≥ 2mn + 2np + 2pm.
From AM-GM it follows that m2n2+ m + n ≥ 3mn ≥ 2mn, from where it is easy to conclude the result.
Second solution by Nguyen Manh Dung, HUS, Hanoi, Vietnam By expanding, we have
LHS : a2b2c2+ x2b2c2+ y2c2a2+ z2a2b2+ a2y2z2+ b2z2x2+ c2x2y2+ x2y2z2, RHS : x2y2z2+ 2xyz(abz + bcx + cay − ayz − bzx − cxy).
The inequality becomes
a2b2c2+x2b2c2+y2c2a2+z2a2b2+2xyz(ayz +bzx+cxy) ≥ 2xyz(abz +bcx+cay) By the AM-GM inequality, we have
x2b2c2+ xyz · bzx + xyz · cxy ≥ 3xyz · xbc. Adding two similar inequalities, we obtain
x2b2c2+ y2c2a2+ z2a2b2+ 2xyz(ayz + bzx + cxy) ≥ 2xyz(abz + bcx + cay), and we are done. Equality holds if and only if a = b = c = x = y = z = 0. Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; Nguyen Manh Dung, Hanoi, Vietnam; John T. Robinson, Yorktown Heights, NY, USA; Perfetti Paolo, Dipartimento di matematica Universita degli studi di Tor Ver-gata, Italy; Salem Malikic Sarajevo, Bosnia and Herzegovina.
O82. Let ABCD be a cyclic quadrilateral inscribed in the circle C(O, R) and let E be the intersection of its diagonals. Suppose P is the point inside ABCD such that triangle ABP is directly similar to triangle CDP . Prove that OP ⊥ P E. Proposed by Alex Anderson, New Trier High School, Winnetka, USA
Solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain
Since P AP C = P BP D = CDAB, point P is the intersection of two distinct Apollonius’ circles constructed by taking the ratios of the distances to A and C for one, B and D for the other. If AB = CD, the circles degenerate to straight lines that meet only at one point. Otherwise, assuming wlog that AB < CD, then the centers of both circles are on the rays CA, DB from C, B, but not on segments CA or DB respectively. Since both points where the circles meet are symmetric with respect to the line joining the centers of both circles, and this line is outside ABCD, then both points cannot be in ABCD simultaneously, and P is therefore unique.
If AB k CD, then ABCD is an isosceles trapezoid, and P = E, or line P E cannot be defined. Assume henceforth then that AB and CD are not parallel, and call F = AB∩CD. Assume furthermore wlog that BC < DA (if BC = DA then ABCD would be an isosceles trapezium and AB k CD, which we are assuming not to be true). Obviously, line EF contains all points Q such that the distances from Q to lines AB and CD, respectively d(Q, AB) and d(Q, CD), satisfy d(Q,AB)d(Q,CD) = ABCD, since it contains E and passes through the intersection of both lines, and trivially AEB and DEC are similar, or the altitudes from E to AB and CD are proportional to the lengths of the sides AB and CD. Therefore, since AP B and CP D are similar, the altitudes from D to AB and CD are also proportional to AB and CD, or P ∈ EF .
We will now show that P is the point the circumcircles of ABE and CDE, and line EF , meet. Call first P the second point where the circumcircle of ABE and line EF meet. The power of F with respect to the circumcircle of ABE (which is also the power of F with respect to the circumcircle of ABCD is then F E · F P = F A · F B = F C · F D. Therefore, CDP E is also cyclic, and P is also on the circumcircle of CDE. Now, since ABEP and CDP E are cyclic, then ∠P AB = ∠BEF = ∠P ED = ∠P CD, and similarly ∠ABP = ∠AEP = ∠CEF = ∠CDP , or indeed P AB and P CD are similar. Note finally that, if the circumcircles of ABE and CDE where tangent, then ∠ABE = ∠BEF = π − ∠DEF = ∠DCE = ∠ABE, and ABE and CDE are isosceles and similar, or AB k CD.
Call now A0, B0, C0, D0 the second points where P D, P C, P B, P A meet the cir-cumcircle of ABCD. Trivially, ∠ACB0 = ∠ECP = ∠EDP = ∠BDA0, or AB0 = BA0, and similarly AC0 = CA0, AD0 = DA0, or AA0BB0, AA0CC0 and AA0D0D are isosceles trapezii, and AA0 k BB0 k CC0 k DD0. Trivially,
the diagonals AD0 and DA0 of AA0D0D meet at P , which is then in the com-mon perpendicular bisector of AA0, BB0, CC0, DD0, which trivially passes also through O, and A0B0C0D0is the result of taking the reflection of ABCD with re-spect to OP . Therefore, OP is the internal bisector of angles ∠AP A0, ∠BP B0, ∠CP C0 and DP D0. Now, ∠BP E = ∠BAE = ∠CDE = ∠CP E, and P E is the internal bisector of angle ∠BP C = π − ∠BP B0, or P E is the external bi-sector of angles ∠BP B0 and ∠CP C0, and hence perpendicular to their internal bisectors, ie, to OP . The proof is completed.
Remark. Note that this solution includes also the way to construct point P , i.e., the second point where the circumcircles of ABE and CDE meet. If both circles are tangent, then as shown P = E, and ABCD is an isosceles trapezium with AB k CD.
O83. Let P (x) = a0xn+ a1xn−1+ . . . + an, an 6= 0, be a polynomial with complex
coefficients such that there is an m with am an > n m .
Prove that the polynomial P has at least a zero with the absolute value less than 1.
Proposed by Titu Andreescu, University of Texas at Dallas, USA
First solution by Perfetti Paolo, Dipartimento di Matematica Tor Vergata Roma, Italy
If x1, x2, . . . , xn are the roots of P (x) = 0, by the Vi`ete’s formulae
am a0 = (−1)mXx1x2. . . xm, an a0 = (−1)nx1x2. . . xn hence X 1 |x1| |x2| . . . |xn−m| ≥ X 1 x1x2. . . xn−m = am an > n m If ε = min1≤k≤n{|xi|} we have 1 εn−m n n − m ≥X 1 |x1| |x2| . . . |xn−m| > n m
and then ε < 1. The proof is completed.
Second solution by G.R.A.20 Math Problems Group, Roma, Italy The zeros of the polynomial
Q(x) = anxn+ an−1xn−1+ · · · + a0
are {1/wk, k = 1, . . . , n} (note that wk6= 0 because an6= 0).
By Vieta’s formula am an = X I∈In−m Y k∈I 1 |wk|
where In−m is the set of all subsets of {1, 2, . . . , n} such that |In−m| = n − m.
If all zeros of P has the absolute value greater or equal than 1 then 1/|wk| ≤ 1
and for any integer m ∈ [0, n − 1] am an ≤ X I∈In−m 1 = n n − m = n m
and this contradicts the hypothesis.
Third solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain Denote by r1, r2, . . . , rnzeros of P (x), and assume without out loss of generality
that a0= 1, since we may divide all coefficients of P (x) by a0 without changing
the conditions of the problem. For the value of m such that am an > n m, call s1, s2, . . . , s(n
m) the products of each subset of exactly m roots in some order.
Clearly, am is the sum of the sk, and using the triangular inequality,
|an| < P( n m) k=1sk n m ≤ P( n m) k=1|sk| n m ,
and without loss of generality max |sk| = |s1| > |an|, where again without loss
of generality s1 = r1r2· · · rm. Therefore,
|rm+1rm+2· srn| =
|r1r2. . . rn|
|s1| < 1, and at least one of |rm+1|, |rm+2|, . . . , |rn| is less than 1.
O84. Let ABCD be a cyclic quadrilateral and let P be the intersection of its di-agonals. Consider the angle bisectors of the angles ∠AP B, ∠BP C, ∠CP D, ∠DP A. They intersect the sides AB, BC, CD, DA at Pab, Pbc, Pcd, Pda,
respec-tively and the extensions of the same sides at Qab, Qbc, Qcd, Qda, respectively.
Prove that the midpoints of PabQab, PbcQbc, PcdQcd, PdaQda are collinear.
Proposed by Mihai Miculita, Oradea, Romania
Solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain
The internal bisector of angles ∠AP B and ∠CP D is the same straight line, which passes through P, Pab, Pcd, Qbc, Qda. Clearly, the internal bisector of
∠BP C and ∠DP A passes through P, Pbc, Pda, Qab, Qcd. Trivially, both internal
bisectors meet perpendicularly at P since ∠AP B + ∠BP C = π. Therefore, tri-angles PabP Qab, PbcP Qbc, PcdP Qcd and PdaP Qda are right triangles, and the
respective midpoints Oab, Obc, Ocd, Oda of PabQab, PbcQbc, PcdQcd, PdaQda, are
their respective circumcenters.
Now, without loss of generality ∠BP Qab= π−∠AP B2 and ∠AP Qab= π+∠AP B2 ,
or
AQabsin ∠AQabP
AP = sin ∠AP Qab = sin ∠BP Qab=
BQabsin ∠BQabP BP , and AQab BQab = AP BP = APab
BPab, and the circumcircle of PabP Qab is defined as the
Apollonius circle such that BXAX = BPAP.
Note that, denoting by rab the circumradius of PabP Qab, we find
OabA − rab rab− OabB = PabA PabB = QabA QabB = OabA + rab OabB + rab ,
leading to OabA · OabB = r2ab, and A is the result of performing the inversion
of B with respect to the circumcircle of PabP Qab.
Denote by O and R the circumcenter and thecircumradius of ABCD. Then the power of Oab with respect to the circumcircle of ABCD is OOab2 − R2 =
OabA·OabB = rab2 , and the circumcircles of ABCD and PabP Qabare orthogonal.
In an entirely analogous manner, we deduce that the circumcircle of ABCD is orthogonal to the circumcircles of PabP Qab, PbcP Qbc, PcdP Qcd and PdaP Qda.
For each pair of these circles, O is then in their radical axis, and since each pair of circles meet at P , they either meet at a second point P0 also on OP , or they are pairwise tangent at P . In the first case, the centers Oab, Obc, Ocd, Oda
of the four circles are on the perpendicular bisector of P P0, and in the second case, on the perpendicular to OP through P . The conlusion follows.
